The probability that project a will be complete at time t after it begins is given as:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\[ p_a(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_1$} \\

\frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2$} \\

1 & \mbox{if t_2 \leq t}

\right. \][/tex]

In other words, the project requires at least [itex]t_1[/itex] to complete and will take no more than [itex]t_2[/itex]. The probability of completion at any time between [itex]t_1[/itex] and [itex]t_2[/itex] is a linear function of t.

The probability that project b will be complete at time t after it begins is given as:

[tex]

\[ p_b(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_3$} \\

\frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4$} \\

1 & \mbox{if t_4 \leq t}

\right. \][/tex]

Project b will begin the moment project a is complete. Given t, what is the probability [itex]p_{ab}(t)[/itex] that both projects will be complete at time t. Obviously,

[tex]

\[ p_{ab}(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_1 + t_3$} \\

1 & \mbox{if t_2 + t_4 \leq t}

\right. \][/tex].

What is [itex]p_{ab}(t)[/itex] for [itex]t_1 + t_3 \leq t \leq t_2 + t_4[/itex]?

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# Scheduling problem

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